Question

==>100 pts<==

for all x^2 , x^2 + 2ax +(10-3a) >0 , then the interval in which 'a' lies is ?​​

Answer 1

Ax² + bx + c >0 for all values of x only if

a>0 and D<0

Find or calculate discriminate:-

Use this D= b² - 4a×c

x² -4(10-3a) < 0

4(a² +3a -10) <0

(a+5)(a-2) <0

So a belongs to (-5,2)

Hope it helps you ❤

Answer 2

Answer:

As we know, ax2+bx+c>0 for all x ∈ R, iff

a > 0 and D < 0

Given equation is x2+2ax+(10−3a)>0, ∀ x ∈R

Now, D < 0

⇒4a2−4 (10−3a)<0

⇒4 (a2+3a−10)<0

⇒ (a+5) (a+2)<0

⇒a ∈(−5,2)